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(D) The kinetic energy $E$ of a particle in uniform circular motion is given by $E = \frac{1}{2} L \omega$,where $\omega = 2\pi n$ and $n$ is the freq

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$L/4$

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(D) The kinetic energy $E$ of a particle in uniform circular motion is given by $E = \frac{1}{2} L \omega$,where $\omega = 2\pi n$ and $n$ is the frequency of motion.Thus,$E = \frac{1}{2} L (2\pi n) = \pi L n$.This implies $E \propto L 	imes n$,or $L \propto \frac{E}{n}$.Given the initial state $(E_1, n_1, L_1)$ and final state $(E_2, n_2, L_2)$ where $E_2 = \frac{E_1}{2}$ and $n_2 = 2n_1$,we have:$\frac{L_2}{L_1} = \frac{E_2}{E_1} 	imes \frac{n_1}{n_2} = \frac{E_1/2}{E_1} 	imes \frac{n_1}{2n_1} = \frac{1}{2} 	imes \frac{1}{2} = \frac{1}{4}$.Therefore,$L_2 = \frac{L_1}{4} = \frac{L}{4}$.

$A$ particle is performing uniform circular motion with angular momentum $L$. If the frequency of motion is doubled and the kinetic energy is halved,the new angular momentum will be

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